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ki77a [65]
3 years ago
14

How does pH affect enzyme-catalyzed reactions? How does affect enzyme-catalyzed reactions? Energy stored in protons is used to d

rive endergonic reactions. The concentration of protons affects an enzyme's folded structure and reactivity. Proton concentration increases the kinetic energy of the reactants, enabling them to reach their transition state. Protons serve as substrates for most reactions.
Chemistry
1 answer:
Zolol [24]3 years ago
7 0

Answer:

The concentration of protons affects an enzyme's folded structure and reactivity.

Explanation:

Enzymes act within narrow pH limits (optimal reaction pH). Since most enzymes have a protein structure, the variation in pH or temperature affects their enzymatic activity.

To catalyze a reaction, an enzyme binds to one or more reagent molecules. These molecules are the substrates of the enzyme.

In some reactions, a substrate breaks into several products. In others, two substrates join together to create a larger molecule or to exchange parts. In fact, for any biological reaction that can occur to you, there is probably an enzyme to accelerate it.

The part of the enzyme where the substrate binds is called the active site.

The amino acid residues of the active site often have acidic or basic properties that are important for catalysis. Changes in pH can affect these residues and make binding with the substrate difficult.

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Which statement is true of the carbon atoms that make up a diamond
Lunna [17]
Crystal<span> of tetrahedrally bonded carbon atoms in a </span>covalent network lattice<span> witch  </span>crystallizes<span> into the diamond </span>lattice.

6 0
3 years ago
Find the initial concentration of the weak acid or base in each of the following aqueous solutions: (a) a solution of HClO with
Luda [366]

Answer:

a) 0.021 M

b) 0.019 M

Explanation:

To do this, you need to calculate the concentration of ions in solution with the given value of pH for each solution, then, write the chemical equation for both solutions, Set an ICE chart, use the value of Ka and Kb reported for both solutions, and solve for the initial concentration.

This is the general procedure to do it, now let's do it by parts.

<em><u>a) Concentration of HClO pH = 4.6</u></em>

With the given pH, we use the following expression:

pH = -log[H₃O⁺]      From here, we solve for [H₃O⁺]

[H₃O⁺] = 10^(-pH)   (1)

Let's calculate first the hydronium concentration:

[H₃O⁺] = 10^(-4.6) = 2.51x10⁻⁵ M

This value indicates the equilibrium concentration of this ion in solution. Now, to know the initial concentration of the acid, we need to do an ICE chart and write the chemical equation. This is an acid - base reaction, so we need the value of Ka of the acid.

         HClO + H₂O <---------> H₃O⁺ + ClO⁻       Ka = 3x10⁻⁸

I:            Y                                 0          0

C:          -x                                +x         +x

E:           Y - x                            x          x

With this chart, we need to write the expression for Ka which is:

Ka = [H₃O⁺] * [ClO⁻] / [HClO] = x² / Y-x

But we already know the concentration of [H₃O⁺], which is the same for [ClO⁻], and the value of Ka, so all we have to do is replace the values in the above expression and solve for Y:

3x10⁻⁸ = (2.51x10⁻⁵)² / Y - 2.51x10⁻⁵

We can round to Y because "x" is a very small value as it's value of Ka so:

3x10⁻⁸ = (2.51x10⁻⁵)²/Y

Y = (2.51x10⁻⁵)²/3x10⁻⁸

<h2><em>Y = [HClO] = 0.021 M</em></h2>

<em>And this is the initial concentration of the acid.</em>

<u><em>b) Solution of hidrazine pH = 10.2</em></u>

We do the same procedure as part a) with the difference that instead of using Ka , we use Kb and concentration of [OH⁻]. The Kb for hydrazine is 1.3x10⁻⁶

Let's calculate the [OH⁻]:

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

[OH⁻] = 10^(-3.8) = 1.58x10⁻⁴ M

The chemical equation:

          N₂H₄ + H₂O <---------> N₂H₅⁺ + OH⁻    Kb = 1.3x10⁻⁶

I:            Y                                  0           0

C:          -x                                +x           +x

E:         Y-x                                 x           x

Kb = x²/(Y-x)

1.3x10⁻⁶ = (1.58x10⁻⁴)²/Y

Y = (1.58x10⁻⁴)²/1.3x10⁻⁶

<h2><em><u>Y = [OH⁻] = 0.019 M</u></em></h2>

And this is the initial concentration of hydrazine

4 0
3 years ago
A chef finds a sealed container consisting of an ingredient that goes into his restaurant’s secret sauce. The ingredient’s molec
ozzi

The ingredient's molecules would be more free and move faster, then the phase would change to liquid from solid.

<u>Explanation</u>:

  • This task seeks to examine the knowledge of the kinetic phase and characteristics of matters.
  • Here we see the chef described that the ingredient's molecules are moving in place – solid.
  • On the advent of causing the ingredient to change phase by transferring energy into it – the ingredient's molecules would be more free and move faster, then the phase would change to liquid from solid.

6 0
3 years ago
Reference the following attachment
dem82 [27]

The number of molecules of gases in each of the 3 vessels is the same.

<h3>Number of moles</h3>

The first vessel contains 2g of H_2 gas:

Number of molecules in 2 g = 2/2 = 1 mole

The second vessel contains 32 g of O_2 gas:

Number of molecules = 32/32 = 1 mole

The third vessel contains 44 g of CO_2 gas:

Number of molecules = 44/44 = 1 mole

Thus, the 3 vessels contain the same number of moles of gases.

More on the number of moles can be found here: brainly.com/question/12513822

#SPj1

3 0
2 years ago
If a large marshmallow has a volume of 2.50 i n 3 and density of 0.242 g/c m 3 , how much would it weigh in grams? 1 i n 3 =16.3
Illusion [34]

Density of a substance is defined as the mass of the substance divided by the volume.

Density of the substance= 0.242 g cm⁻³  

volume of the substance= 2.50 in³  

As, 1 in³= 16.39 cm³  

So, 2.50 in³= 16.39× 2.50 cm³=40.97 cm³

As ,  

Density=\frac{mass}{volume}  

Mass=volume ×Density

Mass=40.97 × 0.242

Mass=9.916 g.

4 0
3 years ago
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